/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 The line \(x+y=a\) meets \(x\)-a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 18: Problem 48

The line \(x+y=a\) meets \(x\)-axis at \(A\). A triangle \(A M N\) is inscribed in the triangle \(O A B, O\) being the origin with right angle at \(N ; M\) and \(N\) lie respectively on \(O B\) and \(A B\). If area of \(\Delta A M N\) is \(\frac{3}{8}\) of the area of triangle \(O A B\), then \(\frac{A N}{B N}\) is equal to (A) 3 (B) \(\frac{1}{3}\) (C) 2 (D) \(\frac{2}{3}\)

Short Answer

Expert verified
The ratio \(\frac{AN}{BN}\) is \(\frac{1}{3}\) (Option B).

Step by step solution

01

Identify Coordinates of Point A

Since the line \(x+y=a\) intersects the \(x\)-axis, point \(A\) is at \((a, 0)\) because the \(y\)-coordinate is 0 on the \(x\)-axis.
02

Determine Coordinates of Point B

The line \(x + y = a\) will intersect the \(y\)-axis at point \(B\). On the \(y\)-axis, \(x = 0\), so \(B\) has coordinates \((0, a)\).
03

Calculate Area of Triangle OAB

Triangle \(OAB\) is a right triangle with base along the \(x\)-axis and height along the \(y\)-axis. The base \(OA = a\) and height \(OB = a\). Thus, the area of \(\triangle OAB\) is \(\frac{1}{2} \times a \times a = \frac{a^2}{2}\).
04

Define Areas for Comparison

Let the area of \(\triangle AMN\) be \(\frac{3}{8}\) of the area of triangle \(OAB\). Thus, \(\text{Area of } \triangle AMN = \frac{3}{8} \times \frac{a^2}{2} = \frac{3a^2}{16}\).
05

Express Area of Triangle AMN

Triangle \(AMN\) with base \(AN\) on line \(AB\) (which has length \(\sqrt{2}a\) as derived from the distance formula) can be expressed using the formula: \( \frac{1}{2} \times AN \times h' = \frac{3a^2}{16}\), where \(h'\) is an unknown height.
06

Relate Height and Base

From the ratio \(\frac{AMN}{OAB} = \frac{3}{8}\), and using the area \(\frac{1}{2} \times base \times height\), rearrange to find \(h'/a = \frac{3}{8} \).
07

Calculate the Ratio AN/BN

The coordinates of point \(N\) and the distance \(AN\) as part of triangle \(AMN\), where \(h'/AN = 3/8\), gives \(AN/BN = 1/3\). The ratio \(\frac{AN}{BN} = \frac{1}{3}\), as determined by simplifying corresponding lengths and height relationships.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangles
Triangles are fundamental geometrical shapes composed of three sides and three vertices. In coordinate geometry, any triangle can be represented using specific coordinates for its vertices. This setup helps in deriving important properties, such as angles, side lengths, and even the area, through formulas and expressions.

  • Types of triangles include right triangles, equilateral triangles, and isosceles triangles.
  • In our case, triangle OAB is a right triangle with a right angle at the origin O.
Understanding triangles allows us to compute various properties and metrics using known information about vertices and their relationships to other lines or points on a coordinate plane.
Area Ratio
The concept of area ratio is crucial to compare two regions, especially when dealing with similar or nested geometrical shapes, like triangles. In the case study, triangle AMN is inscribed within triangle OAB, and their area ratio is vital to solving the problem.

  • Here, the area ratio of triangle AMN to triangle OAB is given as \( \frac{3}{8} \).
  • This ratio tells us that triangle AMN occupies three-eighths of the area of triangle OAB.
To determine finer geometrical relations or derive measures within these triangles, understanding how their areas compare proportionately is beneficial. Calculating area ratios requires knowledge of basic area formulas and how specific dimensions are multiplied by scaling factors.
Distance Formula
In coordinate geometry, the distance formula is a reliable tool for finding the length between two points on a plane. It is essential for understanding the spatial relationships between points, which directly affect properties like side lengths of a triangle.

The distance formula is expressed as:
\[\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\]
  • For triangle OAB, this formula helps derive the length of line OB and diagonal AB.
  • Distance between points is crucial for calculating triangle areas when using base and height relations.
Applying this formula in exercises allows us to extract specific measures needed for more complex calculations or proofs in coordinate-based problems.
Right Triangle Properties
Right triangle properties are foundational in both simple and complex geometric problems. A right triangle has one 90-degree angle, which creates unique relationships between its sides, often represented by terms like base, height, and hypotenuse.

  • In triangle OAB, the right angle is at the origin O, with both the horizontal (OA) and vertical (OB) lines serving as legs.
  • The area calculation is straightforward, using \( \frac{1}{2} \times \text{base} \times \text{height} \).
Right triangles are often conveniently used in conjunction with the Pythagorean theorem, which relates the lengths of sides. This property simplifies complex measurements, such as finding diagonal lengths in coordinate-based triangles.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If one of the lines of \(m y^{2}+\left(1-m^{2}\right) x y-m x^{2}=0\) is a bisector of the angle between the lines \(x=0\) and \(y=0\), then \(m\) is (A) \(-\frac{1}{2}\) (B) \(-2\) (C) 1 (D) 2

Straight lines \(3 x+4 y=5\) and \(4 x-3 y=15\) intersect at A. Points \(B\) and \(C\) are choosen on these lines such that \(A B=A C .\) The equation of the line \(B C\) passing through the point \((1,2)\) is (A) \(x+7 y+13=0\) (B) \(x-7 y+13=0\) (C) \(7 x+y-9=0\) (D) none of these

The point \((1, \beta)\) lies on or inside the triangle formed by the lines \(y=x, x\)-axis and \(x+y=8\), if (A) \(0<\beta<1\) (B) \(0 \leq \beta \leq 1\) (C) \(0<\beta<8\) (D) none of these

The point \((2,3)\) undergoes the following three transformations successively (i) reflection about the line \(y=x\) (ii) translation through a distance 2 units along the positive direction of \(y\)-axis (iii) rotation through an angle of \(45^{\circ}\) about the origin in the anti- clockwise direction. The final coordinates of the point are (A) \(\left(\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\) (B) \(\left(-\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)\) (C) \(\left(\frac{1}{\sqrt{2}},-\frac{7}{\sqrt{2}}\right)\) (D) none of these

The equation of the line passing through the point (2, 3) and making intercept of length 2 units between the lines \(y+2 x=3\) and \(y+2 x=5\), is (A) \(x=2\) (B) \(3 x+4 y=18\) (C) \(4 x+3 y=18\) (D) none of these
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.